I am new to this topic and I can't find a way to solve the questions even though I identify the forces which do the work....I am more confused at the Kinetic energy part. Can you tell me how to identify it when using work energy in such cases where it is not specified? For example :

If a body of mass m is being hauled up the hill by a force tangent to the trajectory find the work done by this force if coefficient of friction is μ. The acceleration is zero for this case.

My efforts:

I can identify only four forces acting on body :

(1) applied force(F)

(2) normal force (N)

(3) friction (f)

(4) weight(W)

Using Work energy theorem: $$\sum_{\mathrm{all~forces}}W = \Delta K$$ $$W_{W}+W_{N}+W_{F}+W_{f}=\Delta K$$

How will we calculate $\Delta K$ and will work done by friction be constant because the path doesn't seem linear to me?

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  • $\begingroup$ I believe we are supposed to assume that the box reaches the top and stays still up there. $\Delta K$ will then be zero between start and end points. $\endgroup$ – Steeven Jan 19 '18 at 11:29
  • $\begingroup$ @SteevenI can think it to be the only logical one.I hate when we have to assume such things.Also do we have to integrate friction or somethin doesn't seem linear to me?g in this case bc The path $\endgroup$ – user181139 Jan 19 '18 at 11:39
  • $\begingroup$ No, you don't need to integrate. Hint: if part of the trajectory is vertical, what would be the work against friction there? $\endgroup$ – npojo Jan 19 '18 at 12:15
  • $\begingroup$ @npojo I am confused at what you are saying ...can you explain more of that in an answer please $\endgroup$ – user181139 Jan 19 '18 at 12:19
  • $\begingroup$ The question (as highlighted by the cream background) cannot be answered. We need to know if the body is being hauled at constant speed, or at constant acceleration (and if so, how much) … If the hauling were at constant speed, kinetic energy wouldn't come into the reckoning. $\endgroup$ – Philip Wood Jan 19 '18 at 12:35

The most important detail in this question is that the acceleration is zero, which means that the kinetic energy of the block is the same throughout the process. So, this removes the $\Delta KE$ term for us.

$\therefore W_{F}+W_{f}+W_{gravity}=0=\Delta KE$,

I'm just gonna talk about the friction part of the question.

Usually, we consider the length of the path covered by a body to find the work by friction but something else is happening here. You have been provided with the net horizontal displacement of the block, which should serve as a hint that it has some purpose in the question.

I cannot be too explicit with the answer but I would suggest you to consider finding the work by friction for an infinitesimal displacement along the curve. You will have to do a simple manipulation in the equation and you will reach your answer.

  • $\begingroup$ why do we have to use straight trajectory at an angle theta for inegeration instead of the whole trajectory $\endgroup$ – user181139 Jan 19 '18 at 13:12
  • $\begingroup$ Well, you are asking the definition of integration. You know how to find the work, its the dot product between the force and the displacement vector. So if we consider such a infinitesimally small displacement for which we can calculate the work without any sweat, we can get the whole work. So what we do is add all the infinitesimal work for all the infinitesimal displacements along the trajectory. This way we will get the total work by friction. And that's also how integration works. $\endgroup$ – Mitchell Jan 19 '18 at 13:16
  • $\begingroup$ The reason why we are calculating this way is simply because the normal reaction between the block and the plane is changing for every infinitesimal displacement, which gives us a variable force because frictional force depends on the normal force between two relatively sliding bodies. Thats why we have to use integration. Therefore, we consider the case of infinitesimal displacement for which frictional force is constant and we add all such infinitesimal work terms to get the whole work. $\endgroup$ – Mitchell Jan 19 '18 at 13:22
  • $\begingroup$ It's giving μmgl after I did integrationso it's right $\endgroup$ – user181139 Jan 19 '18 at 13:32
  • 1
    $\begingroup$ We don't have to look too much into that because it is given in the question that acceleration is zero, period. However, the way in which this is possible is that the net force on the particle in zero for every infinitesimal change. This can happen if the resultant of F and friction and it's resultant with the normal force exactly cancels the gravitational force. We have 3 variable forces and it not pretty hard to imagine that their net resultant with gravity could be zero. $\endgroup$ – Mitchell Jan 19 '18 at 13:59

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